On the Characters and the Plancherel Formula of Nilpotent Groups ...

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304 DUDLEY If 11 * I/ is a measurable pseudo-norm on H*, then L is defined by a countably additive probability measure on the completion of H* for I/ * 11.l At the moment the converse seems to be an open question. Suppose (xt , t E S) is a sample-continuous Gaussian process over a compact metric space (S, d). Then t -+ xt is continuous from S into H, and e(s, t) = (E(xs - xJ2)l12 defines a pseudo-metric e on S which is continuous for d and hence defines a weaker topology. If (S, e) is Hausdorff, i.e., if x, # xt for s # t, then the d and e topologies on S are equal, and hence the range of the process in H is a GC-set; its closed convex symmetric hull is a GB-set, which, by Theorem 4.7, is not much worse. 5. SEQUENCES OF VOLUMES We shall need the volumes of certain simple sets in Rk. First, suppose A is a simplex, i.e., a convex hull of (h + 1) points x,, ,..., xk , having an interior. Let F be a face of A, i.e., a convex hull of K of its vertices. Let h or h, be Lebesgue measure on Rk and S or Sk-r the (K - 1)-dimensional Lebesgue “surface” or “area” measure. Then h(A) = S(F) d/k, where d is the distance from the vertex not in F to the hyperplane through F. Now suppose x0 = 0 and let di be the distance from xj to the linear span of x0 ,..., xi-r , j = l,..., k. Then h(A) = (fJ dj)/k!. i=l Now, recalling the definitions of V,(C) and EV(C) given is Section 1, we have the following fact (a stronger statement is given as Propo- sition 5.10 below): LEMMA 5.0. If C is a convex set in H and EV(C) < - 1, then C is totally bounded. 1 For the proof, see L. Gross, Abstract Wiener spaces, in Proceedings of the Fifth Berkeley Symposium cm Mathematical Statistics and Probability (1964). University of California Press, Berkeley, 1967.
SIZES OF COMPACT SUBSETS OF HILBERT SPACE 305 Proof. If C is not totally bounded we make the same construction as in the proof of Proposition 2.4. Then for some E > 0, V,(C) is greater than or equal to the volume of the convex hull of 0, fi ,..., f, , so VJC) > E”/tz! for all n. By Stirling’s formula, this contradicts the hypothesis. Q.E.D. Next, let ck = X,(B) w h ere B is a ball of radius 1 in Rk. Then it can be shown by induction that, for any positive integer k, c&+1 = 22”+wk!/(2k + l)!, c2k = m”jk!. Thus by Stirling’s formula we have the following estimate: y+i Cj(Trj)“” (j/274’2 = 1. (5-l) We shall also need the following fact. Let {a,} be a sequence of positive real numbers such that a, JO as n + co. For such a sequence and E > 0 we define n(c) = n({%} , c) = max (?z : a, >, E), h = h({u,}) = infict:f (5p 0 such that Pr(L(C)y>O. C may be replaced in the above inequality by any orthogonal pro- jection P(C), according to Proposition 4.1. Multiplying C by a
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